python利用openpyxl拆分多个工作表的工作簿的方法
人气:0实现按目录拆分工作簿,源数据如下图
按目录拆分成N个文件。
上代码,没有找是否有整个sheet 复制的,先逐个cell复制解决问题。:
# encoding: utf-8 """ @author: 陈年椰子 @contact: hndm@qq.com @version: 1.0 @file: Split_Xls.py @time: 2019/9/24 0028 15:04 说明 """ def Split_Xls(xls_file): from openpyxl import load_workbook from openpyxl import Workbook wb = load_workbook(xls_file) sheet_list = wb.sheetnames print(sheet_list) a_sheet = wb['目录'] for i in range(3,6): sheet_name = a_sheet['B{}'.format(i)].value if sheet_name is None: break if sheet_name == '': break sr_sheet = wb[sheet_name] new_file_name = "{}.xlsx".format(sheet_name) print(sheet_name) wb_tg = Workbook() ws = wb_tg.active ws.title = sheet_name # 两个for循环遍历整个excel的单元格内容 for i, row in enumerate(sr_sheet.iter_rows()): for j, cell in enumerate(row): # print(i,j,cell.value) ws.cell(row=i + 1, column=j + 1, value=cell.value) wb_tg.save(new_file_name) wb_tg.close() wb.close() def Split_Xls2(xls_file): # 这个是通过删除其他的工作表,只留下要保存的工作表,这样就可以整个表复制,包括样式,过程曲折,但能达到效果。 from openpyxl import load_workbook wb = load_workbook(xls_file) sheet_list = wb.sheetnames print(sheet_list) work_list = [] a_sheet = wb['目录'] for i in range(3,6): sheet_name = a_sheet['B{}'.format(i)].value if sheet_name is None: break if sheet_name == '': break work_list.append(sheet_name) wb.close() for sheet_name in work_list: new_file_name = "{}.xlsx".format(sheet_name) print('处理工作表', sheet_name, '\t保存文件', new_file_name) wb = load_workbook(xls_file) # print(wb.sheetnames) for del_sheet in sheet_list: if del_sheet != sheet_name: # print('del',del_sheet) wb.remove(wb[del_sheet]) wb.save(new_file_name) wb.close() Split_Xls2('test.xlsx')
您可能感兴趣的文章:
加载全部内容